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Suppose $X$ has a continuous, strictly increasing cdf $F$. Let $Y = F(X)$. What is the density of $Y$? Then let $U \sim unif(0,1)$ and let $X = F^{-1}(U)$. Show that $X\sim F$.

The first part seems like the density of $Y$ is just the pdf $f(X)$. The wording of the problem is confusing me though.

Craig
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    What does $X \sim F$ even mean? – Clarinetist Apr 04 '15 at 19:33
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    I wish I knew... that's a huge part of why the wording/set up is confusing me so much. It doesn't seem to make sense to say that $X$ is distributed as $F$, a cdf. I was hoping there was another possible interpretation that I was missing. – Craig Apr 04 '15 at 19:53
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    The notation $X\sim F$ might be slightly on the lax side but it is completely standard. Your comment sounds as if you had no textbook or lecture notes at your disposal and were trying to rediscover the subject through guesses. Is that so? – Did Apr 04 '15 at 20:13
  • @Did - What does $X \sim F$ even mean in this context? – Clarinetist Apr 04 '15 at 20:16
  • ^ That $F$ is the distribution function of $X$, i.e. $\mathbb P(X\leqslant x)=F(x)$ for $x\in\mathbb R$. – Math1000 Apr 04 '15 at 20:18
  • @Clarinetist Hint: $X$ is a random variable, $F$ is a CDF, so what do you think $X\sim F$ could mean? – Did Apr 04 '15 at 20:18
  • @Did - Oh, I misread. For some reason I thought it was trying to make a circular argument based on assuming it had the cdf $F$ on the first sentence. – Clarinetist Apr 04 '15 at 20:22
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    This is a classical problem. (The answer is used in computer simulation.) To start look at 1197188 on this site (beginning of answer). – BruceET Apr 04 '15 at 20:32
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    Also, maybe look at 1208396 – BruceET Apr 04 '15 at 20:38

1 Answers1

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Since $F$ is continuous and monotone increasing, it has an inverse function $F^{-1}$, i.e. for any $t\in\mathbb R$, there is a unique number $F^{-1}(t)$ such that $F^{-1}(F(t)) = t$. Recall that the inverse of a monotone increasing function is also monotone increasing. We have $$\mathbb P(Y\leqslant t) = \mathbb P(F(X)\leqslant t). $$ Since $F$ takes values in $[0,1]$, it is clear that $\mathbb P(Y\leqslant t)=0$ for $t\leqslant 0$ and $\mathbb P(Y\leqslant t)=1$ for $t\geqslant 1$. For $t\in(0,1)$ we have $$\mathbb P(F(X)\leqslant t)=\mathbb P(F^{-1}(F(X))\leqslant F^{-1}(t)) = \mathbb P(X\leqslant F^{-1}(t)) = F(F^{-1}(t)) = t. $$ It follows that $$\mathbb P(Y\leqslant t)=t 1_{(0,1)}(t) + 1_{[1,\infty)}(t), $$ and therefore the density $g$ of $Y$ is the derivative of the above: $$g(y) = 1_{(0,1)}(t). $$ In other words, $Y$ has $\mathcal U(0,1)$ distribution.

For the second part, we have for any $t\in\mathbb R$ $$ \begin{align*} \mathbb P(X\leqslant t) &= \mathbb P(F^{-1}(U)\leqslant t)\\ &= \mathbb P(F(F^{-1}(U))\leqslant F(t))\\&=\mathbb P(U\leqslant F(t))\\&=\mathbb P(U\leqslant t)\\&= F(t). \end{align*} $$

Math1000
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  • Why if F is monotonic increasing and continous, that implies that F has inverse? For example, consider F=1 on all $\mathbb{R}$ which is monotonic (and not stricly) increasing and is continuous, however F has no inverse in $\mathbb{R}$ – HeMan Jan 10 '16 at 10:14
  • In this problem, we assumed that $F$ was strictly increasing, so $x<y$ implies $F(x)<F(y)$. – Math1000 Jan 10 '16 at 11:55